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Let ABC be a triangle with AB=3cm, AC=5cm. If AD is a median drawn from the vertex A to the side BC, then which one of the following is correct?

a) AD is always greater than 4cm but less than 5cm

b) AD is always greater than 5cm

c) AD is always less than 4cm

d) none of the above

I assumed the triangle to be right angled at B and got BC=4. So, BD=2. I used Median theorem, AB^2 + AC^2 = 2(AD^2 + BD^2), and got AD=sqrt(13), which is less than 4. So, I am left with c) and d) options.

How to move forward?

And is there a way to solve this question instinctively without actually finding the value of AD?

aarbee
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    I think it would be worth looking at the possible values for side BC. We know that 2 < BC < 8 so from there you can look at what happens in a few cases using AB^2 + AC^2 = 2(AD^2 + BD^2), and from this you should be able to determine the answer. – KBusc Sep 05 '14 at 14:26

3 Answers3

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Since the maximum that side BC can be is 8 then using the Median theorem we get

$9 + 25 = 2AD^2 + 2(8/2)^2 ==> 9 + 25 = 2AD^2 + 32$

So, $2 = 2AD^2 ==> 1 = AD^2$ so AD can be as small as 1

Now if we look at the minimum that BC can be we get

$9 + 25 = 2AD^2$ + $2(2/1)^2$ ==> $9 + 25 = 2AD^2 + 2$

So, $32 = 2AD^2$ ==> $16 = AD^2$ and AD = 4

So the max AD can be is 4 and the min Ad can be is 1 so we have

$1 \leq AD \leq 4$.

So the answer would be C

KBusc
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  • That's a very nice explanation. Thanks. But I wish you help me in the second part as well, that is, finding answer without finding the value of AD. Is that possible? – aarbee Sep 05 '14 at 15:01
  • I guess I am unsure of what you mena. In this answer and in Jack D'Aurizio's the value of AD is never found. It is just shown to be less than 4 in all cases. – KBusc Sep 05 '14 at 15:09
  • @Ramit after looking at this again since 4 is a possible answer for the median it leads me to believe that the correct answer might actually be D – KBusc Sep 05 '14 at 15:12
  • @Ramit it looks like C actualy is the answer. take a look at the comments under Jack's answer to see why. – KBusc Sep 05 '14 at 15:21
  • @KBusc- Thanks. – aarbee Sep 05 '14 at 15:25
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Stewart's theorem gives that the length of the median $m_a$ between the two sides $b$ and $c$ satisfies: $$ 4m_a^2 = 2b^2+2c^2-a^2 \tag{1}$$ and since $a\geq |b-c|$ by the triangle inequality, we have: $$ 4m_a^2 \leq (b+c)^2 \tag{2}$$ hence the length of the median $m_a$ is always less or equal to the arithmetic mean of $b$ and $c$.

On the other hand, since $a\leq b+c$, $$ 4m_a^2 \geq (b-c)^2.\tag{3}$$ In you case, $(2)$ and $(3)$ give: $$1\leq m_a \leq 4 $$ hence the right answer is c).

Jack D'Aurizio
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    That's some answer! I must note it down. Thanks. – aarbee Sep 05 '14 at 15:05
  • @Jack D'Aurizio as I was looking at your answer and mine I realized that since 4 is a value for the median are we wrong in saying C is the correct answer, since C says Always less than 4? Should the answer be D? – KBusc Sep 05 '14 at 15:11
  • @KBusc: It is kind of subtle, since $m_a=4$ only if $a=b+c$, i.e. only when $ABC$ is a degenerate triangle (side lengths $3,5,2$). For "usual" triangles, the inequality holds as $m_a < 4$, hence c) is right, indeed. – Jack D'Aurizio Sep 05 '14 at 15:17
  • @JackD'Aurizio that was my original thought. In my original answer I had the inequality with the strictly less than symbol because thats how I have always used it. I was under the impression that a 3,5,2 triangle was not actually consider to be a triangle. – KBusc Sep 05 '14 at 15:20
  • @KBusc: I agree with you and I think that $3,5,2$ is not considered as a triangle in this case, so c) applies with a strict inequality. – Jack D'Aurizio Sep 05 '14 at 15:23
  • Ain't it simply called Apollonius Theorem? – Sawarnik Oct 23 '14 at 06:39
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Think about the two degenerate cases, one where $B$ lies between $A$ and $C$, and the other where $A$ lies between $B$ and $C$, and then about how the median point moves when $B$ does.

rogerl
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